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Home/ Questions/Q 8254237
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Editorial Team
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Editorial Team
Asked: June 8, 20262026-06-08T01:07:21+00:00 2026-06-08T01:07:21+00:00

A = [a1, a2, a3…] #a1<a2<a3… B = [b1, b2…] #b1<b2<b3… A and B

  • 0
A = [a1, a2, a3...]  #a1<a2<a3...
B = [b1, b2...]  #b1<b2<b3...

A and B are disjoint. are I do not know the number of elements and the value of them in A/B in advance. I want to compare the value of the elements in both list and delete elements iff:

delete a[i+1] if there is no b[j] such that a[i]<b[j]<a[i+1]
delete b[i+1] if there is no a[j] such that b[i]<a[j]<b[i+1]

At the end, I want to separate list, not a combination of A and B.

For example, If A[0] < B[0], A = [1, 10, 40], B = [15, 30]. Compare A[1] and B[0] first. Delete 10 because no element in B are in between 1 and 15.
Then delete 15 since no element exist anymore btw 15 and 30. The output should be: if you try to order the elements of the new 2 lists, it should be A[0]<B[0]<A[1]<B[1]<...

If A[0] > B[0], vice versa.

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  1. Editorial Team
    Editorial Team
    2026-06-08T01:07:23+00:00Added an answer on June 8, 2026 at 1:07 am
    a = [1, 10, 40]
    b = [15, 30]
    
    srcs = [a, b]
    dsts = [[], []]
    prev_which = -1
    while all(srcs):
        which = int(srcs[0][0] > srcs[1][0])
        elem = srcs[which].pop(0)
        if prev_which != which:
            dsts[which].append(elem)
        prev_which = which
    for src, dst in zip(srcs,dsts):
        if src:
            dst.append(src.pop(0))
    a, b = dsts
    

    returns:

    a = [1, 40]
    b = [15]
    

    and for

    a = [3, 4, 6, 7, 8, 9]
    b = [1, 2, 5, 10]
    

    it returns [3, 6] and [1, 5, 10].

    EDIT: another possibility:

    import itertools as it
    import operator as op
    
    a = [3, 4, 6, 7, 8, 9]
    b = [1, 2, 5, 10]
    srcs = [a, b]
    dsts = [[], []]
    
    for which, elems in it.groupby(sorted((x, i) for i in (0,1) for x in srcs[i]), key=op.itemgetter(1)):
        dsts[which].append(next(elems)[0])
    a, b = dsts
    
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